## Calculus (3rd Edition)

$$y'= \frac{-19}{(4t-9)^2}.$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $y=\frac{3t-2}{4t-9}$, then using the quotient rule, the derivative $y'$ is given by $$y'= \frac{(4t-9)(3)-(3t-2)(4)}{(4t-9)^2}= \frac{-19}{(4t-9)^2}.$$